AMC 8 · 2012 · #5
Grade 4 geometry-2dProblem
In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , X in centimeters?
Pick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A closed shape has only right angles, so every edge is either horizontal or vertical. All the side lengths are labeled in centimeters except one vertical edge marked $X$. The figure is not to scale, so we must trust the labels, not the picture. Find $X$.
Givens: All interior angles are right angles, so every side is either horizontal or vertical; Labeled vertical lengths (in centimeters): $6, 1, 2, 1, 1, 1, 1, 2$; Labeled horizontal lengths (in centimeters): $1, 2, 3, 2, 1, 4, 2, 2$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: The length $X$ of the unlabeled vertical edge, in centimeters
Understand
Restated: A closed shape has only right angles, so every edge is either horizontal or vertical. All the side lengths are labeled in centimeters except one vertical edge marked $X$. The figure is not to scale, so we must trust the labels, not the picture. Find $X$.
Givens: All interior angles are right angles, so every side is either horizontal or vertical; Labeled vertical lengths (in centimeters): $6, 1, 2, 1, 1, 1, 1, 2$; Labeled horizontal lengths (in centimeters): $1, 2, 3, 2, 1, 4, 2, 2$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #2 Make a Systematic List
Because the figure closes up, if we walk all the way around the perimeter we must end up exactly where we started. That means the total upward distance must equal the total downward distance (otherwise we would float above or sink below the start). Tool #1 (Draw a Diagram) says: copy the figure and mark each vertical edge with an up-arrow or a down-arrow as you trace the boundary. Tool #2 (Make a Systematic List) says: write the up-lengths in one column and the down-lengths in another, then add each column. Setting up-total $=$ down-total gives a single equation for $X$ — no algebra harder than basic subtraction.
Execute — Answer: E
3.G.A.1 Step 1 - Pick a starting vertex and walk around the boundary once, marking each vertical edge with an arrow showing whether you move up or down on that edge.
- Horizontal edges do not affect $X$, so we ignore them for now.
💡 Recognizing that a closed rectilinear polygon is made only of horizontal and vertical edges is a Grade 3 shape-attribute idea.
3.MD.D.8 Step 2 - List every upward edge with its label.
- Tracing the boundary, the four edges that go up are the ones labeled $6$, $1$ (just above the $1$-step jog on the right), $2$ (far right), and $1$ (the small step near the top right).
💡 Sorting edges into "up" and "down" piles is the systematic-list move; we are organizing the perimeter data.
4.OA.A.3 Step 3 Add the up-edge lengths to get the total upward distance traveled around the loop.
💡 Adding labeled whole-number lengths is exactly the Grade 4 multi-step word-problem skill.
3.MD.D.8 Step 4 - Now list every downward edge with its label.
- Tracing onward, the downward edges are labeled $1$ (top-left jog), $1$ (just below it), $1$ (far left), $2$ (a longer drop on the left), and finally $X$ (the unknown drop back to the start).
💡 Same systematic-list move, applied to the opposite direction so nothing is double-counted.
4.OA.A.3 Step 5 - Add the down-edge lengths.
- The known downs sum to $5$, so the full downward total is $5 + X$.
💡 Combining labeled lengths into a single expression sets up the closure equation.
3.MD.D.8 Step 6 - Because the loop closes, total up must equal total down.
- Set the two sums equal and find $X$ by subtracting the known downs from $10$.
💡 Finding a missing side length from the rest of the boundary is exactly the Grade 3 perimeter standard, just applied direction-by-direction.
3.G.A.1 Pick a starting vertex and walk around the boundary once, marking each vertical 3.MD.D.8 List every upward edge with its label. Tracing the boundary, the four edges that 4.OA.A.3 Add the up-edge lengths to get the total upward distance traveled around the loo 3.MD.D.8 Now list every downward edge with its label. Tracing onward, the downward edges 4.OA.A.3 Add the down-edge lengths. The known downs sum to $5$, so the full downward tota 3.MD.D.8 Because the loop closes, total up must equal total down. Set the two sums equal Review
Reasonableness: Quick horizontal sanity check on the same closed-loop idea: right-going edges are $3, 1, 2, 2, 2 = 10$ and left-going edges are $3, 2, 1, 4 = 10$. They match, so the figure really does close, which means our vertical equation is being read the right way. The answer $X = 5$ is one of the offered choices and the largest, which fits the picture where $X$ is one of the longer vertical drops.
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: try $X = 5$ first. Then the downward total is $1 + 1 + 1 + 2 + 5 = 10$, which equals the upward total $6 + 1 + 2 + 1 = 10$. The loop closes, so (E) works. Any smaller choice ($1, 2, 3, 4$) gives a downward total less than $10$, so the boundary would not close — those choices are all eliminated.
CCSS standards used (min grade 4)
3.G.A.1Understand that shapes in different categories may share attributes (e.g., having right angles) (Recognizing the figure as a rectilinear polygon — every edge is either horizontal or vertical — so the boundary splits cleanly into up, down, left, and right moves.)3.MD.D.8Solve real-world and mathematical problems involving perimeters of polygons, including finding an unknown side length (Using the rest of the labeled edges to recover the unknown side length $X$ — exactly the "missing side from the perimeter" task, applied direction-by-direction.)4.OA.A.3Solve multistep word problems posed with whole numbers using the four operations (Adding the up-edge lengths ($6 + 1 + 2 + 1 = 10$), adding the known down-edge lengths ($1 + 1 + 1 + 2 = 5$), then subtracting $10 - 5 = 5$ to isolate $X$.)
⭐ This AMC 8 problem only needs Grade 4 thinking: if a shape closes up, the total "up" must equal the total "down" — so the missing side is whatever balances the two columns.
⭐ This AMC 8 problem only needs Grade 4 thinking: if a shape closes up, the total "up" must equal the total "down" — so the missing side is whatever balances the two columns.
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